Question

1.

Given the following contingency table, conduct a test for independence at the 1% significance level. (You may find it useful to reference the appropriate table: chi-square table or F table)

	Variable A
Variable B		1		2
1		31		32
2		34		58

Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

2.

A market researcher for an automobile company suspects differences in preferred color between male and female buyers. Advertisements targeted to different groups should take such differences into account if they exist. The researcher examines the most recent sales information of a particular car that comes in three colors. (You may find it useful to reference the appropriate table: chi-square table or F table)

	Sex of Automobile Buyer
Color	Male	Female
Silver	477	298
Black	536	308
Red	482	348

Calculate the value of the test statistic. (Round the intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

3.

Consider the following sample data with mean and standard deviation of 20.1 and 7.3, respectively. (You may find it useful to reference the appropriate table: chi-square table or F table)

Class		Frequency
Less than 10			27
10 up to 20			80
20 up to 30			60
30 or more			21
		n = 188

Calculate the value of the test statistic. (Round the z value to 2 decimal places, all other intermediate values to at least 4 decimal places and final answer to 3 decimal places.)

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H₀: Variable A and variable B are independent.
H_a: Variable A and variable B are not independent.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a chi-square test for independence.

Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (2 - 1) * (2 - 1)
D.F = 1
E_r,c = (n_r * n_c) / n

Χ² = 2.30

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 1 degrees of freedom is more extreme than 2.30.

We use the Chi-Square Distribution Calculator to find P(Χ² > 2.30) = 0.131

Interpret results. Since the P-value (0.131) is greater than the significance level (0.01), we have to accept the null hypothesis.

Thus, we conclude that there is no relationship between variable A and variable B.